m(ar AB) = 60°, m(ar BAC) = 240°, m(ar BC) = 120°,
m(ar EA) = 92°, m(ar ECA) = 268°
Solution:
Let us take O be the center of the circle.
Given m∠AOB = 60° and m∠BOE = 32°
Sum of the adjacent angles = 180°
m∠EOC + 32° + 60° = 180°
m∠EOC + 92° = 180°
m∠EOC = 88°
Angle in the diameter is 180°.
∠AOC = 180°
The angle measure of the central angle is congruent to the measure of the intercepted arc.
m∠AOB = m(ar AB) = 60°
Central angle = intercepted arc
m∠AOC = m(ar AOC) = 180°
m(ar BAC) = m(ar AOC) + m(ar AOB)
= 180° + 60°
m(ar BAC) = 240°
m(ar BC) = m(ar BE) + m(ar EC)
= 32° + 88°
m(ar BC) = 120°
m(ar EA) = m(ar EB) + m(ar AB)
= 32° + 60°
m(ar EA) = 92°
m(ar ECA) = m(ar EC) + m(ar AOC)
= 88° + 180°
m(ar ECA) = 268°
